# (When) does a morphism of monad induce adjoint functors between categories of algebras?

For monads $$S$$ and $$T$$ on a fixed Abelian category $$C$$, a morphism of monads $$\sigma: S\rightarrow T$$ induces a functor between Eilenberg-Moore categories $$\sigma^*:C^T\rightarrow C^S$$. This functor sends a $$T$$-algebra $$(A,\rho:TA\rightarrow A)$$ to an $$S$$-algebra $$(A,\rho\circ \sigma_A:SA\rightarrow A)$$. This functor commutes with forgetful functors $$U^T:C^T\rightarrow C$$ and $$U^S:C^S\rightarrow C$$.

Are there any known results that would also provide a left adjoint $$\sigma_!$$ to $$\sigma^*$$? In addition, I would like this left adjoint to commute with the free algebra functors for $$T$$ and $$S$$, i.e. $$F^T=\sigma_!\circ F^S$$.

Does it even make sense to expect such a functor to exist?

The reason I'm looking for a left adjoint is that I want to work with a category of Abelian categories whose morphisms are pairs of adjoint functors. Monads $$S$$ and $$T$$ are induced by an adjoint pair, there is a canonical $$free \dashv forgetful$$ adjunction for EM categories, so I would like to have an adjoint for $$\sigma^*$$ too.

First note that the "in addition" part is automatic because left adjoints compose and by design $$U^S\sigma^* = U^T$$.

Second, it is not unreasonable to expect $$\sigma_!$$ to exist : $$\sigma^*$$ preserves all limits that exist in $$C$$, so under common set-theoretic assumtpions, $$\sigma_!$$ will exist for free. But in this specific situation, one can be more explicit and relax the assumptions: as I said (and as you required), the value of $$\sigma_!$$ on free algebras is determined, and free algebras "generate" all algebras.

In particular, if $$C^T$$ admits reflexive coequalizers, then $$\sigma_!$$ exists.

These will exist if, e.g. $$C$$ has them (it does if $$C$$ is abelian as in your question) and $$T$$ preserves them (all monads preserve a certain amount of reflexive coequalizers, but not necessarily all). In that case, $$U^T : C^T\to C$$ in fact creates reflexive coequalizers, but this need not happen in general, even if $$C^T$$ has them.

In a different direction, let me spell out the set theory: if $$C$$ is presentable and $$T$$ is accessible, then $$C^T$$ is presentable and thus admits reflexive coequalizers (although in this case, the adjoint functor theorem also kicks in).

• In the last paragraph: It also works if $C$ is the opposite of a locally presentable category. Apr 16 at 19:01

To complement the answer of Maxime with a useful reference, my go-to article for this circle of questions is "Coequalizers in categories of algebras" by Fred Linton (LNM 80, Seminar on Triples and Categorical Homology Theory, ETH 1966/67).